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Proof step 3 from proof in-prepend-other (previous step)

⊦ (∃ eqent. eqent.pnum = n ∧ (eqent = eq-entry(n0, lref) ∨ eqent ∈ q)) ↔ (∃ eqent. eqent.pnum = n ∧ eqent ∈ q), n = n0

Heuristic pl case distinction applied rule case distinction

And obtained the following new goals

Proof step 4:
∃ eqent. eqent.pnum = n ∧ (eqent = eq-entry(n0, lref) ∨ eqent ∈ q) ⊦ n = n0, ∃ eqent. eqent.pnum = n ∧ eqent ∈ q
Proof step 8:
∃ eqent. eqent.pnum = n ∧ eqent ∈ q ⊦ n = n0, ∃ eqent. eqent.pnum = n ∧ (eqent = eq-entry(n0, lref) ∨ eqent ∈ q)

Rule argument

pos: right 1

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